Question

Prove that 5√3 is irrational.
explanation needed.

Answer 1

Your answer is...

It's exactly the same as proving 2–√2is irrational.

Suppose 5=(ab)35=(ab)3 where a,ba,b are integers and gcd(a,b)=1)gcd(a,b)=1) [i.e. the fraction is in lowest terms].

The 5b3=a35b3=a3 so 5 divides a3a3 but as 5 is prime (indivisible) it follows 5 divides aa. So a=5a′a=5a′ for some integer a′a′.

So 5b3=(5a′)3=125a′35b3=(5a′)3=125a′3 so b3=25a′3b3=25a′3.

So 2525 divides b3b3. So 55 divides b3b3. But as 5 is prime (indivisible) it follows that 5 divides bb.

So you have 55 divides aa AND we have 55 divides bb. But that contradicts gcd(a, b) = 1 and that the fraction is in lowest terms.

We must conclude such a fraction is impossible and 5–√353 is irrational
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Answer 2

assume that 5√3 is rational

⇒ 5√3 = a/b , where a and b are integers .

⇒ √3= a/5b

we know that a, b and 5 are integers and they are also rational {i.e RHS is rational}

therefore root 3 will be rational.
but we know that √3 is irrational.
there is a contradiction

so, 5√3 is an irrational number
∴ it is proved that 5√3 is irrational.

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